Question

Nuclei of a radioactive `A` are being produceted at a constant rate a . The electro has a decay constant `lambda` . At time `t = 0 ` there are `N_(0) `nucleius of the element .
(a) calculate the number `N` of nuclei of `A` at same t
(b) if a = 2N_(0) lambda) ` calculate the number of nuclei of `A` after one half - life of `A` and the limiting value of `N` as `t rarr oo`

Answer 1

Correct Answer - A::B::C::D

Let at time `t` number of radioactive nuclie are `N` net rate of formation of nuclie of `A`
`(dN)/(dt) = a - lambda, N or (dN)/(a - lambda N) = dt `
`or int _(N_(0)^(N) (dN)/(a - lambda N ) = int _(0)^(1) dt `
Solving this equation , we get
`N = (1)/(lambda) [a - (a - lambda N_(0)) e^(- lambda t)]`
Subsituting `a = 2 lambda n_(0) `and
`t = t_(1//2) = (in (2))/(lambda) ` in equation (1),
we get , `N = (2)/(3) N_(0)`
(ii) subsitituting `a = 2 lambda N_(0) and t rarr oo ` equation (1) we get
` N = (a)/( lambda) = 2 N _(0)`